Simplifying rational powers after completing a derivative

[andraos]

I have the following function to differentiate:

f(x) = $$x^\frac{4}{5} * (x-4)^2$$

My derivative is:

f'(x) = $$x^\frac{4}{5} * 2(x-4) + (x-4)^2 * (\frac{4}{5}) * x^\frac{-1}{5}$$

My calculator gives me:

f'(x) = $$\frac{2(x-4) * (7x-8)}{5x^\frac{1}{5}}$$

How does my expression simplify into the latter?

It must have something to do the (x-4) term and x^m * x^n = x^(m+n) but I can't quiet see it.

Thanks for the help,
Salim

A step-wise explanation would be awesome!

 

[Mark44]

Your derivative expression is a sum of terms that involve x - 4 and x to different powers. The general idea is to find the smaller power of each term and factor it out.

For example, if you had (x + 2)³y² + (x + 2)y³, both terms involve powers of x + 2 and y. Of the two terms, the smallest power of x + 2 is the 1st power, and the smallest power of y is the 2nd power, so you can factor this expression in this way.

(x + 2)³y² + (x + 2)y³
= (x + 2)y²( (x + 2)² + y²)

All I've done is use the distributive property to write ab + ac as a(b + c).

In your problem the smallest power of x - 4 is the 1st power, and the smallest power of x is the (-1/5) power, so we can factor these quantities out to get

(x - 4)x^-1/5(2x¹ + (4/5)(x - 4))

= 2x^-1/5(x - 4)(5x/5 + 2x/5 - 8/5)

= 2x^-1/5(x - 4)(1/5)(7x - 8)

$$= \frac{2(x - 4)(7x - 8)}{5x^{1/5}}$$

 

[andraos]

Thanks a lot Mark. Excellent explanation!